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I think the answer is 10.

previous answer and explanation:

1st batter reaches, steals 2nd and 3rd, gets thrown out at home. (2 SB, 1 out) 2nd batter reaches, steals 2nd and 3rd, gets thrown out at home. (4 SB, 2 out) 3rd batter reaches, steals 2nd and 3rd. (6SB, 2 out) 4th batter reaches and steals 2nd. (7 SB, 2 out)

There were 0 runs scored.

my answer and explanation:

I think the answer is 10. Two runners steal 2nd and 3rd and get picked off for 4 stolen bases and two outs. The third batter also steals 2nd and 3rd to bring the total to 6. The 4th batter walks and steals 2nd for the 7th stolen base. Everything to this point was how the previous answer of 7 was discussed. Then the 5th hitter walks. Then all three runners try to steal. The runner from third gets caught in a rundown. The runner from 2nd steals 3rd, and the runner from first steals 2nd and 3rd to bring the total to 10 stolen bases before any of the 3 runners now between 3rd and home get tagged out. Also the 3rd out occurs before home plate is touched by the lead runner.

Read more: What_is_the_highest_amount_of_stolen_bases_in_one_inning

Question:

I don't think the runners from 1st or 2nd can be awarded steals until the completion of a play, which would be when the runner from third is tagged, thus ending the inning. This is a weird scenario, but when there is a double steal attempt and the runner is thrown out at 3rd base for the last out of the inning, the runner going to 2 does not get credit for a steal, even though he may have reached the bag prior to the tag.

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12y ago

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